Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 22989

Homotopy theory, homological algebra, algebraic treatments of manifolds.

6 votes

free action on contractible spaces

The example I gave in comments has $E/G_1$ and $E/G_2$ compact (but not manifolds), at least if you use compact "blobs". More explicitly, let $E$ be the subset of $\mathbb{R}^2$ given by the union of …
Jeremy Rickard's user avatar
12 votes

Is every paracompact, Hausdorff, locally contractible space homotopy equivalent to a CW comp...

I think Milnor answers your question in the paper where he proves the theorem you refer to ("On spaces having the homotopy type of a CW-complex", Trans. AMS, 90 (1959), 272-280): Whitehead had observ …
Jeremy Rickard's user avatar
2 votes

Technical but elementary homotopy question

Can't you just take $X=[0,1]$ and $K=\{0,1\}$, with $$f(x,t)=\begin{cases}\frac{x}{1-t}&\mbox{if $t\leq 1-x$}\\ 1&\mbox{if $t\geq 1-x$},\end{cases}$$ so that $f(x,t)\to 1$ as $t\to 1$ unless $x=0$, in …
Jeremy Rickard's user avatar
15 votes
Accepted

Which spaces have the (weak) homotopy type of compact Hausdorff spaces?

Expanding on my comment, if there are measurable cardinals then it follows from the results of A. Przeździecki, Measurable cardinals and fundamental groups of compact spaces. Fund. Math. 192 (2006), …
Jeremy Rickard's user avatar
6 votes

If an abelian category $\mathcal{A}$ has enough injectives then so does $\mathrm{Ch}^{\geq 0...

If $X^{\bullet}$ is an object of $\mathrm{Ch}^{\geq0}(\mathcal{A})$ and for each $i\geq0$ $X^i\to I^i$ is an embedding into an injective, then $X^{\bullet}$ embeds in $$I^{\bullet}:=\dots\to 0\to I^0\ …
Jeremy Rickard's user avatar
10 votes
Accepted

Group cohomology with coefficients in a permutation module

I'm not sure I understand your intuition, but the statement is true. It's a special case of Shapiro's Lemma: if $H$ is a subgroup of $G$, $A$ is a $\mathbb{Z}H$-module, and $A^G$ is the coinduced modu …
Jeremy Rickard's user avatar
6 votes

A sort of "group-ring" construction on coefficient systems in group homology (+ special case...

$R[M]$ is just a permutation module, so, by Shapiro's Lemma, $H_{\ast}(G;R[M])$ is the direct sum, over a set of representatives $m$ of orbits of $G$ acting on $M$, of the homology $H_{\ast}(G_m;R)$ w …
Jeremy Rickard's user avatar
4 votes
Accepted

When are automorphisms in categories homotopically trivial?

How about adding another object $W$ to your example with an automorphism $\delta:W\to W$ with $\delta^{-1}=\delta$ and arrows $\theta,\phi:X\to W$ with $\delta\circ\theta=\phi=\theta\circ\gamma$? The …
Jeremy Rickard's user avatar
5 votes

Free and cellular G-action implies free G-complex?

If $G$ acts freely on a CW-complex, permuting the cells, then the stabilizer of a cell must be finite (and therefore trivial, as pointed out in the question). This can be shown by induction on the di …
Jeremy Rickard's user avatar
13 votes
Accepted

Z/p action on finite contractible complex

If a finite $p$-group $P$ acts on a finite-dimensional $CW$-complex $X$ which is acyclic mod $p$, then the fixed point set $X^P$ is also acyclic mod $p$. This is a special case of "Smith theory" (see …
Jeremy Rickard's user avatar
20 votes
Accepted

Is the homotopy category of an abelian model category abelian?

No. The projective model structure on chain complexes of modules over a ring is an abelian model category, and the homotopy category is the derived category, which is never abelian unless the ring is …
Jeremy Rickard's user avatar
6 votes

Unbounded acyclic resolutions

I'm afraid this is not very close to the case that you say you're most interested in ... maybe you want $F$ to preserve products? But let $A=k[x]/(x^2)$, and let $\mathscr{A}$ be the category $\operat …
Jeremy Rickard's user avatar
4 votes

Homological vs. cohomological dimension of a group/space

As noted in the question, "Question 3" reduces to asking whether, if $H_i(G,M)=0$ for all $i>0$ and all $\mathbb{Z}G$-modules $M$, then $G$ must be trivial. But this is true, since if $G$ is non-triv …
Jeremy Rickard's user avatar
30 votes

If $A$, $B$ are abelian groups such that $\mathrm{Hom}(A, G) \cong \mathrm{Hom}(B, G)$ for a...

I just stumbled across the answer to this in Fuchs' 2015 book on Abelian Groups. The papers Hill, Paul, Two problems of Fuchs concerning tor and hom, J. Algebra 19, 379-383 (1971). ZBL0228.20027. and …
Jeremy Rickard's user avatar